User alexander alldridge - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T11:10:03Z http://mathoverflow.net/feeds/user/17229 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/6922/references-for-harish-chandra-pairs-and-modules-category-o/91986#91986 Answer by Alexander Alldridge for References for Harish-Chandra pairs and modules, category "O"? Alexander Alldridge 2012-03-23T08:28:31Z 2012-03-23T08:28:31Z <p>This answer may be a little late, considering the date it was asked, but anyway...</p> <p>Concerning question 4), a "standard" reference is perhaps the two volume treatise by Wallach on "Reductive Groups". Some aspects of Harish-Chandra modules (with a view towards globalisation questions) are also developed in a recent preprint by Bernstein and Kroetz (see B. Kroetz's web page). Jantzen's Habilitationsschrift has already been mentioned. If you can read German, this also an excellent source.</p> <p>Over the last few years, Penkov, Serganova and Zuckermann have been developing the theory of (g,k)-modules in the general setting where g is some Lie algebra and k is some subalgebra. The modules one considers are g-modules which are semi-simple over k (plus possibly additional conditions). Check the list of recent preprints and publications of these authors.</p> http://mathoverflow.net/questions/54927/morphisms-between-supermanifolds-r01r01/73150#73150 Answer by Alexander Alldridge for Morphisms between supermanifolds R^{0|1}→R^{0|1} Alexander Alldridge 2011-08-18T13:38:23Z 2011-08-18T13:46:35Z <p>@Ma: As an answer to your following question:</p> <blockquote> <p>Could you give a clue how to calculate the $map$, for example, $map(R^{0∣d},M)$ for any supermanifold $M$?</p> </blockquote> <p>Take a look at arXiv:math/0307303, where this question is discussed. </p> <p>For $d=1$, it is well-known (and due to Kontsevich, I think), that $map(R^{0|1},M)$ is the total space of the odd tangent bundle $\Pi TM$ of $M$. </p> <p>If $U$ is a superdomain of dimension $p|q$, then $map(R^{0|d},U)=U\times R^{pr+qs|ps+qr}$ where $(r+1)|s=2^{d-1}|2^{d-1}$ is the graded dimension of $\bigwedge R^d$. This you can check using the definition of $map$ and the characterisation of morphisms of supermanifolds as given in Leites.</p> <p>Another good source on this subject (for $d=1$), is the paper "Differential forms and 0-dimensional supersymmetric field theories" by Hohnhold, Kreck, Stolz and Teichner.</p> http://mathoverflow.net/questions/65684/gluing-of-manifolds-and-the-hausdorff-condition/73147#73147 Answer by Alexander Alldridge for Gluing of manifolds and the Hausdorff condition. Alexander Alldridge 2011-08-18T13:23:11Z 2011-08-18T13:23:11Z <p>I am leaving this as an answer since I am new here and therefore can't comment on answers. It's not an answer, since I am not saying anything about the nerve of the covering. Apologies to the regulars here!</p> <p>Here's what I wanted to say: David Carchedi's answer to the question is rather nice. However, it can be stated in much simpler terms, as follows. </p> <p>Notice that the condition on the map <code>$\coprod U_{\alpha\beta}\to\coprod U_\alpha\times\coprod U_\beta$</code> to be proper is simply that it be closed, since it is injective and its domain is Hausdorff. Moreover, it's open, so it's a homeomorphism onto its image, so it is closed if and only if its range is closed. The range of the map is just the graph of the equivalence relation $R$ on <code>$X':=\coprod U_\alpha$</code> such that $X'/R$ is the glued space $X$. </p> <p>Thus the statement is that $X$ is Hausdorff if and only $R$ has a closed graph in $X'\times X'$. This is true for any equivalence relation $R$ such that the canonical projection $X'\to X'/R=X$ is open. You may find this in Chapter 1, § 8.3 of Bourbaki, General Topology, vol. 1. The type of equivalence relation $R$ coming from such gluings as considered always has this property. This is also in Bourbaki, somewhere in § 4. Anyway, these things are not difficult to check by hand.</p>